What is a system of equations for the following situation? There is a total of 40 questions on a test. Some questions are multip
1 answer:
First we define the variables:
m = multiple choice questions
s = short answer questions
Next we need to satisfy 2 conditions:
1) There are a total of 40 questions. We can represent this mathematically using the variables I defined:
2) The multiple-choice questions are worth 2 points and the short-answer are worth 4 points but they need to total 100 points. This can also be represented mathematically:
By satisfying the conditions provided we were able to create the 2 equations!
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Answer:
26 cm²
Step-by-step explanation:
The area of the rectangle whose dimensions are shown at the right and bottom is ...
(6 cm)(7 cm) = 42 cm²
The figure is smaller than that by the area of the space whose dimensions are shown at the right and in the middle left:
(4 cm)(4 cm) = 16 cm²
The figure area is then the difference ...
42 cm² - 16 cm² = 26 cm²
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<em>Alternate solution</em>
Draw a diagonal line between the lower right inside corner and the lower right outside corner. This divides the figure into two trapezoids.
The trapezoid at lower left has bases 7 and 4 cm, and height 6-4 = 2 cm. Its area is ...
A = (1/2)(b1 +b2)h = (1/2)(7 + 4)(2) = 11 . . . . cm²
The trapezoid at upper right has bases 6 cm and 4 cm and height 3 cm. Its area is ...
A = (1/2)(b1 +b2)h = (1/2)(6 + 4)(3) = 15 . . . . cm²
Then the area of the figure is the sum of the areas of these trapezoids, so is ...
11 cm² + 15 cm² = 26 cm²
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<em>Comment on other alternate solutions</em>
There are many other ways you can find the area of this figure. It can be divided into rectangles, triangles, or other figures of your choice. The appropriate area formulas should be used, and the resulting partial areas added or subtracted as required.
You can also let a geometry program find the area for you. (It is 26 cm².)
